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# If y is an integer, is y^3 divisible by 12? (1) y is

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If y is an integer, is y^3 divisible by 12? (1) y is [#permalink]

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08 Feb 2009, 13:04
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If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.

(2) y is divisible by 6.

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Joined: 07 Jan 2009
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08 Feb 2009, 13:37
This one is where you have to factor everything out. 12 factors into what? 2,6 -> 2,2,3

1. (x4)^3 -> x,4,4,4 -> 2,2,2,2,2,2,x only if x factors into 3. NO
2. (x6)^3 -> x,6,6,6 -> 2,3,2,3,2,3,x YES it doesn't matter what x is because we have the 2,2,3 needed for 12

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08 Feb 2009, 15:12
IMO B

From stmt 1, it is given y is divisible by 4 ==> y has atleast 2,2 as the factors. $$y^3$$ will have atleast 2 6's. From these, we cannot form 12 for sure. Hence insifficient.

From stmt 2, it is given that y is divisible by 6. ==> y has atleast 2 and 3. Now $$y^3$$ will have 2 3's and 3 3's. From these, we can make 12 as one of the factor. Hence it is sufficient.

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08 Feb 2009, 15:19
lumone wrote:
If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.

(2) y is divisible by 6.

B.

y= 2*3 k
y^3 = 2*3 *2*3*2*3 k^3 -->clearly divisble by 12
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Re: DS- 14   [#permalink] 08 Feb 2009, 15:19
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